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Reply 1

Is it y=log8x or y=log8 * x

Reply 2

Assumming you mean log(base 8):

y = log(base 8) x

=> x = 8^y (*)

1. log(base 8) x^2 = 2.log(base 8) x [using basic log laws]

=> log(base 8) x^2 = 2y

2. From (*):

x = 8^y = (2^3)^y = 2^3y

Taking log(base 2) of each side:

log(base 2) x = 3y

3. Use these values in the equation given:

3.(2y) + 3y = 6
9y = 6
y = 1/3

Plug that back in (*) to get your value of x. Apologies if this is wrong, rushed through it :smile:

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